∵AA',BB'为三角形ABC的外角,角EAB和角CBD的角分线,且AA'=AB=BB'
∴∠EAA'=A'AB,∠CBB'=∠B'BD,∠AA'B=∠ABA',∠BAB'=BB'A
∴∠ACB=∠AB'B+∠CB'B=∠AB'B+1/2∠CBD=∠AB'B+1/2(∠BAC+∠ACB)
=∠BAC+1/2(∠BAC+∠ACB)
∴∠ACB=3∠BAC,·········①
又∠ABA'=∠ACB+∠BAC=180°-2∠A'AB+∠ACB=180°-2(180°-2∠ABA')+∠ACB
∴3∠ABA'+∠ACB=180°
∴3(∠BAC+∠ACB)+∠ACB=180°,··········②
由①②得,∠ACB=36°